Hi, have you tried solving the square binomial and then apply vertex formula ( ) and then substitute y = 0. Finally you can find and

Results 1 to 7 of 7

- April 14th 2013, 04:20 PM #1

- Joined
- Apr 2013
- From
- ohio
- Posts
- 7

## Can someone help me with this ? Thanks so much !

**I need help with some algebra !!**

Here is the problem: find the Vertex of the Parabola y= -2(x-2)^2+10

I need the Vertex. ( I got 1.99?) x= 2.0 ?

Then I have to find the horizontal intercepts of the Parabola where y = 0.

Then I have to put it in factored form of the parabola in Form y = a( x - x1 )( x - x2 ))

- April 14th 2013, 04:37 PM #2

- April 14th 2013, 04:41 PM #3

- Joined
- Apr 2013
- From
- ohio
- Posts
- 7

## Re: Can someone help me with this ? Thanks so much !

I came up with a Vertex of 1.99 rounded to 2.0? But have no idea where or what to do next. I tried using my graphing calculator and entered the equation into y = but it comes up error. I do not know what I am doing wrong...

- April 14th 2013, 04:41 PM #4

- Joined
- Apr 2005
- Posts
- 17,277
- Thanks
- 2135

## Re: Can someone help me with this ? Thanks so much !

How in the world did you get 1.99? If x= 2 then x- 2= 0 so -2(x- 2)^2= 0 and y= 10. For any other x, x- 2 is non-zero so (x- 2)^2 is positive and -2(x- 2)^2 is negative so y is less than 10.

Yes, the "horizontal" or "x" intercepts, where the graph crosses the x-axis, are where y= 0. So you need to solve the equation -2(x- 2)^2+ 10= 0 That is, of course, the same as

-2(x- 2)^2= -10 or, dividing by -2, (x- 2)^2= 5.

Obviously a(x- x1)(x- x2) will be 0 where x= x1 or x= x2 so those have to be the points where y= 0 you just found.

- April 14th 2013, 05:07 PM #5

- Joined
- Apr 2013
- From
- ohio
- Posts
- 7

## Re: Can someone help me with this ? Thanks so much !

Thanks for answering but I still do not understand how to come up with the answer to my problems. Sorry I just don't get this algebra stuff. I have 3 weeks left in my college algebra class before I graduate. Math is by far my worst subject. I was hoping someone could tell me step by step what I need to do ? Cause I'm lost. I appreciate you taking the time to answer my post.

- April 14th 2013, 06:08 PM #6
## Re: Can someone help me with this ? Thanks so much !

Actually there are not many steps to follow and the idea is help you to understand. As I said, have you tried using the vertex formula I wrote?

I guess you have solved this types of equations (or at least have an example to follow), so I suggest you to expand if you're completely lost. That will lead you to the*traditional*form of quadratic functions.

There are a few formulas that are meant to use in order to find some special points from the curve.

Let's say your function looks like: you could:

Find its vertex: and

Find its roots (x values which makes y = 0):

As you can see the symbol indicates that there are two possible values ( and ).

Of course, as said by HallsofIvy, you could find the roots easier by solving for x in the given equation. Anyways, there isn't much more than that.

(Note: it's not by coincidence that the x position of the vertex turned out to be 2. The function given is written in a special form, which I can't recall its name, where the number substracting the x and the independent term are the vertex coordinates x and y repectively)

- April 14th 2013, 06:14 PM #7
## Re: Can someone help me with this ? Thanks so much !

@tracieemra72:

We are not a homework service, we are a*help*service. We are not going to simply solve your problems for you. You need to try to follow what the other members are suggesting and work it out as best you can, then post how far you've got. Please check Forum Rule #5.

-Dan